Question: Divide the following complex numbers. $ \dfrac{-8+24i}{4-4i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4+4i}$ $ \dfrac{-8+24i}{4-4i} = \dfrac{-8+24i}{4-4i} \cdot \dfrac{{4+4i}}{{4+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8+24i) \cdot (4+4i)} {(4-4i) \cdot (4+4i)} = \dfrac{(-8+24i) \cdot (4+4i)} {4^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8+24i) \cdot (4+4i)} {(4)^2 - (-4i)^2} = $ $ \dfrac{(-8+24i) \cdot (4+4i)} {16 + 16} = $ $ \dfrac{(-8+24i) \cdot (4+4i)} {32} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8+24i}) \cdot ({4+4i})} {32} = $ $ \dfrac{{-8} \cdot {4} + {24} \cdot {4 i} + {-8} \cdot {4 i} + {24} \cdot {4 i^2}} {32} $ Evaluate each product of two numbers. $ \dfrac{-32 + 96i - 32i + 96 i^2} {32} $ Finally, simplify the fraction. $ \dfrac{-32 + 96i - 32i - 96} {32} = \dfrac{-128 + 64i} {32} = -4+2i $